# Homework 50

- Page ID
- 28964

**10.38**

The air inside of a blimp is heated to create lift. The blimp as a system expands to do 81 kJ of work, and absorbs 150 J of heat,. What is the change of the internal energy of the system? (Answer to be provided in kJ)

**STEP ONE: **Identify the variables that we are given, which are: Heat (q) and Work (w). Where q=150 (J) and w=81 (kJ)

**STEP TWO: **Identify what we are searching for, which is: Change in Internal Energy \[\Delta E\].

**STEP THREE: **Realize that the equation required for this problem is \[\Delta E=q+w\]

**STEP FOUR: **The problem requests for an answer in (kJ), so we need to convert Heat (q) from (J) to (kJ). This is done through dimensional analysis \[(\frac{150J}{1})*(\frac{1kJ}{1000J})=.15kJ\] So, we have 0.15 (kJ) of Heat (q).

**STEP FIVE: **Now that we have our units correct we must check to see whether or not the Work (w) done is negative or positive. Work (w) is positive when work is done **ON** the system, however Work (w) is negative when work is done **BY** the system. The question defines the system as the blimp, which means that the work must be negative since the work is done **BY **the system as the blimp expands. So Work (w) is **-**81 kJ.

**STEP SIX:** Now that we have correct units and determined the work to be negative we can plug the variables into the equation. \[\Delta E=(.15kJ)+(-81kJ)\] So, \[\Delta E=-80.85kJ\] Thus, the change in internal energy is -80.85kJ

**ANSWER:** -80.85 (kJ)

**11.61**

What is the total pressure in (atm) of a mixture of gases in a closed container with the partial pressures as indicated: H, 115 torr; Ar, 105 torr; and N_{2}, 204 torr? What is the mass of H, Ar, and N_{2} at 28°C and 1.45-L?

**STEP ONE: **Find total pressure of the system by adding all of the partial pressures. \[115 torr +105torr+204torr=total pressure\]\[424torr=total pressure\]

**STEP TWO: **Convert total pressure to atmospheres through dimensional analysis.\[(\frac{424torr}{1})\cdot(\frac{1atm}{760torr})=0.558atm\] this is our total pressure.

**STEP THREE: **Now we must try and find the masses of each of the substances, to do this we must first find the fractional pressures, and convert them to atmospheres by multiplying by the total pressure. \[H=(\frac{115torr}{424torr})=0.2712\] \[Ar=(\frac{105torr}{424torr})=0.2476\] \[N_{2}=(\frac{204torr}{424torr})=0.4811\] Next we multiply by the total pressure to get our fractional pressures in atmospheres. \[H=(0.2712)(0.558atm)=0.151atm\] \[Ar=(0.2476)(0.558atm)=0.138atm\] \[N_{2}=(0.4811)(0.558atm)=0.268atm\]

**STEP FOUR: **Before we plug in our numbers we must find the volume for each element, to do this we multiply our fractional pressures by the total volume. \[H=(0.2712)(1.45L)=0.393L\] \[Ar=(0.2476)(1.45L)=0.359L\] \[N_{2}=(0.4811)(1.45L)=0.698L\]

**STEP FIVE: **The final step before we can use our equation is to convert our temperature from Celsius to Kelvin. \[28°C+273= 301K\]

**STEP FIVE:** Now we must utilize the equation \[PV=nRT, where P=pressure(atm), V=volume(L), n=moles, R=.0821\frac{L*atm}{mol*K}, T=Temperature(K))\] to find mole of the substance. We modify this equation to give us. \[n=\frac{PV}{RT}\]

**STEP SIX: **Now we use this equation to find our moles of each gas. \[n(H)=\frac{(0.151atm)(0.393L)}{(0.0821\frac{L*atm}{mol*K})(301K)}=0.0024mol\] \[n(Ar)=\frac{(0.138atm)(0.359L)}{(0.0821\frac{L*atm}{mol*K})(301K)}=0.0020mol\] \[n(N_{2})=\frac{(0.268atm)(0.698L)}{(0.0821\frac{L*atm}{mol*K})(301K)}=0.0076mol\]

**STEP SEVEN:** Now that we have the moles we must multiply this by the molar mass of each gas to find its value in grams. \[H=(0.0024mol)(1.008\frac{g}{mol})=0.0024g\] \[Ar=(0.0020mol)(39.95\frac{g}{mol})=0.0799g\] \[N_{2}=(0.0076mol)(28.02\frac{g}{mol})=0.2130g\]

**ANSWER: **\[Total Pressure=0.558atm\] \[H=0.0024g\] \[Ar=0.0799g\] \[N_{2}=0.2130g\]